# NCERT solutions for Class 12 Physics Chapter 3 Current Electricity

**NCERT solutions for class 12 physics chapter 3 Current Electricity - **Current is the flow of electric charge through a given area. If we want to utilize the charge flow we must have a closed path. Solutions of NCERT class 12 physics chapter 3 explains questions about the factors affecting the current flow, how current is flowing in a conductor, basic laws related to current electricity and circuit analysis and a few measuring devices. Class 12 Physics chapter 3 NCERT solutions are related to Ohm’s law and Kirchoff's law which will help in solving circuit related problems in exams. The NCERT solutions for class 12 physics chapter 3 also explains problems related to heat developed in a resistor. The basics of these chapters are covered in NCERT of high school classes also.

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Also, check NCERT solutions for class 12 other subjects.

## ** NCERT solutions for class 12 physics chapter 3 Current Electricity Exercise **

### ** Answer ** :

Given, the emf of battery, E = 12 V

Internal resistance of battery, r = 0.4 Ohm

Let I be the maximum current drawn from the battery.

We know, according to Ohm's law

E = Ir

I = E/r = 12/0.4 =30 A

Hence the maximum current drawn from the battery is 30 A.

** Answer: **

Given, The emf of the battery, E = 10 V

The internal resistance of the battery, r = 3 Ohm

Current in the circuit, I = 0.5 A

Let R be the resistance of the resistor.

Therefore, according to Ohm's law:

E = IR' = I(R + r)

10 = 0.5(R + 3)

R = 1 Ohm

Also,

V = IR (Across the resistor)

= 0.5 x 17 = 8.5 V

Hence, terminal voltage across the resistor = 8.5 V

** Answer: **

We know that when resistors are combined in series, the effective resistance is the sum of that resistance.

Hence, total resistance of the three resistance combination = 1 + 2 + 3 = 6 Ohm

** Answer: **

Since the resistances are in series, the current through each one of them will be equal to the current through the circuit but voltage/ potential drop will be different.

Total resistance, R = 6 Ω

Emf, V = 12 V

According to Ohm's law:

V = IR

12 = I x 6

I = 12/6 = 2 A

Now, using the same relation, voltage through resistors:

1 Ω : V(1) = 2 x 1 = 2V

2 Ω : V(2) = 2 x 2 = 4V

3 Ω : V(3) = 2 x 3 = 6V

(Note: V(1) + V(2) + V(3) = 2 + 4 + 6 = 12 V )

** Answer: **

We know that when resistances are in parallel combination, the total resistance R is given by:

Therefore, total resistance of the given three resistances in parallel combination is

Hence, the total resistance is 1.05 Ω

** Answer: **

Since the resistances are in parallel, the voltage across each one of them will be equal.

Emf, V = 20 V

According to Ohm's law:

V = IR I = V/R

Therefore, current across each one of them is:

2 Ω : I = 20/2 = 10 A

4 Ω : I = 20/4 = 5 A

5 Ω : I = 20/5 = 4 A

** Answer: **

Given,

temperature coefficient of filament,

=

;

Let T_{2} be the temperature of element, R_{2}=117 Ω

(Positive alpha means that the resistance increases with temperature. Hence we can deduce that T_{2} will be greater than T_{1} )

We know,

Hence, the temperature of the element is 1027 °C.

** Answer: **

, where is the resistivity of the material

Hence, the resistivity of the material of wire is

** 3.7 ** A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.

** Answer: **

Given,

;

;

We know,

2.7 = 2.1[1 + (100 - 27.5) ]

= (2.7 - 2.1) ** / ** 2.1(100 -27.5)

= 0.0039

Hence, the temperature coefficient of silver wire is 0.0039

** Answer: **

For the given voltage, the two values of current will correspond to two different values of resistance which will correspond to two different temperature.

V = 230 V

and

Using Ohm's law:

and

Now, the temperature coefficient of filament,

=

Let be the steady temperature of the heating element.

We know,

That is

230/2.8 = 230/3.2[1 + ( ) ( T _{ 2 } - 27) ]

T _{ 2 } = (840.5 + 27) °C

Hence, steady temperature of the element is ** 867.5 °C ** .

** 3.9 ** Determine the current in each branch of the network shown in Fig. 3.30:

** Answer: **

Let current in the circuit is distributed like

where I1, I2, and I3 are the different current through shown branches.

Now, applying KVL in Loop

Also, we have

so putting it in kvl equation

.................................(1)

Now let's apply kvl in the loop involving I1 I2 AND I3

.................................(2)

now, the third equation of KVL

..............................(3)

Now we have 3 equation and 3 variable, on solving we get

Now the total current

** Answer: **

Balance point from the end A, l _{ 1 } = 39.5 cm

Resistance of Y = 12.5 Ω

We know, for a meter bridge, balance condition is:

The connections between resistors in a Wheatstone or meter bridge made of thick copper strips to minimise the resistance of the connection which is not accounted for in the bridge formula.

** ** (b) Determine the balance point of the bridge above if X and Y are interchanged.

** Answer: **

If X and Y are interchanged.

Then, X = 12.5 Ω , Y = 8.2 Ω

We know, for a meter bridge, balance condition is:

** Answer: **

if the galvanometer and cell are interchanged the galvanometer will show no current and hence no deflection.

** Answer: **

Emf of battery, E = 8 V

Internal resistance of battery, r = 0.5 Ω

Supply Voltage, V = 120 V

The resistance of the resistor, R = 15.5 Ω

Let V' be the effective voltage in the circuit.

Now, V' = V - E

V' = 120 - 8 = 112 V

Now, current flowing in the circuit is:

I = V' / (R + r)

Now, using Ohm ’s Law:

Voltage across resistor R is v = IR

V = 7 x 15.5 = 108.5 V

Now, the voltage supplied, V = Terminal voltage of battery + V

Terminal voltage of battery = 120 -108.5 = 11.5 V

The purpose of having a series resistor is to limit the current drawn from the supply.

** Answer: **

Let E_{2} be the voltage in the second case.

Now, the balance condition is given by :

Therefore, the emf of the second cell = 2.25 V

** Answer: **

We know,

v_{d} :drift Velocity = length of wire(l) / time taken to cover

by substituting the given values

t = 2.7 x 10^{4} s

Therefore, the time required by an electron to drift from one end of a wire to its other end is s.

## ** NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity Additional Exercises **

** Answer: **

Given, the surface charge density of earth

=

Current over the entire globe = 1800 A

Radius of earth, r = 6.37 x 10^{6} m

The surface area of earth A =

= =

Now, charge on the earth surface,

Therefore,

Let the time taken to neutralize earth surface be t

Current I = q / t

t = 282.78 s.

Therefore, time take to neutralize the Earth's surface is 282.78 s

** Answer: **

Given,

There are 6 secondary cells.

Emf of each cell, E = 2 V (In series)

The internal resistance of each cell, r = 0.015 Ω (In series)

And the resistance of the resistor, R = 8.5 Ω

Let I be the current drawn in the circuit.

I = 1.4 A

Hence current drawn from the supply is 1.4 A

Therefore, terminal voltage, V = IR = 1.4 x 8.5 = 11.9 V

** Answer: **

Given,

Emf , E = 1.9 V

Internal resistance, r =380 Ω

The maximum current that can drawn is I = E/r = 1.9/380 = 0.005 A

The motor requires a large value of current to start and hence this cell cannot be used for a motor of a car.

** Answer: **

We know,

R = Ρ l / A

The wires have the same resistance and also are of the same length.

Hence,

Now, mass = Density x Volume = Density x Area x length

Taking the ratio of their masses for the same length

Hence,

Therefore, for the same resistance and length, the aluminium wire is lighter.

Since aluminium wire is lighter, it is used as power cables.

** 3.17 ** What conclusion can you draw from the following observations on a resistor made of alloy manganin?

Current A | Voltage V | Current A | Voltage V |
---|---|---|---|

0.2 | 3.94 | 3.0 | 59.2 |

0.4 | 7.87 | 4.0 | 78.8 |

0.6 | 11.8 | 5.0 | 98.6 |

0.8 | 15.7 | 6.0 | 118.5 |

1.0 | 19.7 | 7.0 | 138.2 |

2.0 | 39.4 | 8.0 | 158.0 |

** Answer: **

The ratio of Voltage to current for the various values comes out to be nearly constant which is around 19.7.

Hence the resistor made of alloy manganin follows Ohm's law.

Answer the following questions:

** Answer: **

The current flowing through the conductor is constant for a steady current flow.

Also, current density, electric field, and drift speed are inversely proportional to the area of cross-section. Hence, not constant.

Answer the following questions:

** Answer: **

No. Ohm’s law is not universally applicable for all conducting elements.

A semiconductor diode is such an example.

Answer the following questions

** 3.18 (c) ** A low voltage supply from which one needs high currents must have very low internal resistance. Why?

### Answer:

Ohm's law states that: V = I x R

Hence for a low voltage V, resistance R must be very low for a high value of current.

Answer the following questions:

** 3.18 (d) ** A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?

** Answer: **

A very high internal resistance is required for a high tension supply to limit the current drawn for safety purposes.

** 3.19 ** Choose the correct alternative:

** (a) ** Alloys of metals usually have (greater/less) resistivity than that of their constituent metals. ** (b) ** Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals. ** (c) ** The resistivity of the alloy manganin is nearly independent of/ increases rapidly with increase of temperature. ** (d) ** The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of ( ).

** Answer: **

(a) Alloys of metals usually have ** greater ** resistivity than that of their constituent metals.

(b) Alloys usually have much ** lower ** temperature coefficients of resistance than pure metals.

(c) The resistivity of the alloy manganin ** is nearly independent of ** temperature.

(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of 10^{22} .

** 3.20 (a) **

** Answer: **

To get maximum effective resistance, combine them in series. The effective resistance will be nR.

** 3.20 (a) **

** Answer: **

To get minimum effective resistance, combine them in parallel. The effective resistance will be R/n.

** 3.20 (a) **

** (iii) ** What is the ratio of the maximum to minimum resistance?

** Answer: **

The ratio is nR/(R/n) =

** 3.20 (b) **

** Answer: **

We have, equivalent resistance = 11/3

Let's break this algebraically so that we can represent it in terms of 1, 2 and 3

this expression is expressed in terms of 1, 2 and 3. and hence we can make a circuit which consist only of 1 ohm, 2 ohms and 3 ohms and whose equivalent resistance is 11/3. that is :

** 3.20 (b) **

** Answer: **

Connect 2 Ω and 3 Ω resistor in parallel and 1 Ω resistor in series to it

Equivalent Resistance R = {1/(1/2 + 1/3)} + 1 = 6/5 + 1

R = 11/5 Ω

** 3.20 (b) **

** Answer: **

1 Ω+2 Ω+ 3 Ω= 6 Ω, so we will combine the resistance in series.

** 3.20 (b) **

** Answer: **

Connect all three resistors in parallel.

Equivalent resistance is R = 1/(1/1 + 1/2 + 1/3) = (1x 2 x 3)/(6 + 3 + 2)

R = 6/11 Ω

** 3.20 (c) **

** (a) ** Determine the equivalent resistance of networks shown in

Fig. 3.31.

** Answer: **

It can be seen that in every small loop resistor 1 ohm is in series with another 1 ohm resistor and two 2 ohms are also in series and we have 4 loops,

so equivalent resistance of one loop is equal to the parallel combination of 2 ohms and 4 ohm that is

now we have 4 such loops in series so,

Hence equivalent resistance of the circuit is 16/3 ohm.

** 3.20 (c) **

** (b) ** Determine the equivalent resistance of networks shown in Fig. 3.31.

** Answer: **

It can be seen that all 5 resistors are in series, so

Equivalent Resistance = R + R + R + R + R = 5R

Hence equivalent resistance is 5R.

** Answer: **

First, let us find the equivalent of the infinite network,

let equivalent resistance = R'

Here from the figure, We can consider the box as a resistance of R'

Now, we can write,

equivalent resistance = R'

' =[( R')Parallel with (1)] + 1 + 1

Since resistance can never be negative we accept

, We have calculated the equivalent resistance of infinite network,

Now

Total equivalent resistance = internal resistance of battery+ equivalent resistance of the infinite network

= 0.5+1+1.73

=3.23 ohm

Hence current drawn from the 12V battery is 3.72 Ampere.

** Answer: **

Given

maintained constant emf of standard cell = 1.02V, balanced point of this cell = 67.3cm

Now when the standard cell is replaced by another cell with emf = . balanced point for this cell = 82.3cm

Now as we know the relation

hence emf of another cell is 1.247V.

(b) What purpose does the high resistance of 600 k have?

** Answer: **

If a sufficiently high current passes through galvanometer then it can get damaged. So we limit the current by adding a high resistance of 600 k .

(c) Is the balance point affected by this high resistance?

** Answer: **

No, the Balance point is not affected by high resistance. High resistance limits the current to galvanometer wire. The balance point is obtained by moving the joe key on the potentiometer wire and current through potentiometer wire is constant. The balance point is the point when the current through galvanometer becomes zero. The only duty of high resistance is to supply limited constant current to potentiometer wire.

** Answer: **

No, the method would not have worked if the driver cell of the potentiometer had an emf of 1.0V instead of 2, because when emf of the driving point is less than the other cell, their won't be any balance point in the wire

** Answer: **

No, the circuit would not work properly for very low order of Voltage because the balance points would be near point A and there will be more percentage error in measuring it. If we add series resistance with wire AB. It will increase the potential difference of wire AB which will lead to a decrease in percentage error.

** 3.23 ** Figure 3.34 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

** Answer: **

Given,

the balance point of cell in open circuit =

value of external resistance added =

new balance point =

let the internal resistance of the cell be .

Now as we know, in a potentiometer,

hence the internal resistance of the cell will be 1.68

## Importance of solutions of NCERT for class 12 chapter 3 current electricity in board exams:

As far as the CBSE board exam and competitive exams are considered, the solutions of NCERT Class 12 Physics chapter 3 are important. In NEET and JEE Main previous year papers, 7 to 10% questions are asked from this chapter. In CBSE board exam 2019, 6 marks questions are asked from current electricity. Use these CBSE NCERT solutions for class 12 physics chapter 3 PDF for a better score in board and competitive exams.

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## NCERT solutions for class 12 physics chapter wise

### Subject wise solutions

### Last Minute Tips

This NCERT Solutions for Class 12 Physics chapter 3 will help in practicing questions for annual exams.

Students can solve and match their answers from this class 12 Physics chapter 3 NCERT solutions.

These solutions will give proper answers that will help students in exams to score good marks.

## Frequently Asked Question (FAQs) - NCERT solutions for Class 12 Physics Chapter 3 Current Electricity

**Question: **How important is the NCERT Solutions for Class 12 Physics chapter 3 for CBSE board exam?

**Answer: **

For CBSE board exam on an average 8 to 10 % questions can be expected from the chapter current electricity. Certain papers of CBSE asked around 15% questions from NCERT chapter 3 current electricity.

**Question: **What are the important topics of current electricity?

**Answer: **

If we analyse the previous year NEET, JEE Main and board papers we can see that the topics like meter bridge, potentiometer and problems using kcl and kvl and combinations of resistance and cells are frequently asked.

**Question: **What is the weightage of current electricity for JEE main?

**Answer: **

For JEE main 2 to 3 questions can be expected from the chapter current electricity.

**Question: **Whether NCERT is enough to cover current electricity for NEET?

**Answer: **

The concepts in NCERT chapter current electricity is a must for NEET exam. Along with NCERT practising previous year papers and giving mock tests would be enough.

**Question: **What is the weightage of current electricity for NEET exam?

**Answer: **

Current electricity carries 8% of weightage on an average for NEET exam.

**Question: **How important is the NCERT Solutions for Class 12 Physics chapter 3 for higher studies in the field of engineering?

**Answer: **

Current electricity is the basics of electrical and electronics engineering related branches. In these branches analysis and designs of the circuit are important and the basic laws studied in the Class 12 Physics chapter 3 NCERT solutions help for the same.

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